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Pulley On A Slope

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data
    In a stage production of Peter Pan, the 60 kg actress playing Peter has to fly in vertically, and to be in time with the music, she must be lowered, starting from rest, a distance of 3.2 m in 2.2 s. Backstage, a smooth surface sloped at 59º supports a counterweight of mass m.

    What is the mass of the counterweight that must be used? The pulley is of negligible mass and is frictionless. The acceleration of gravity is 9.81 m/s². Answer in units of kg.


    2. Relevant equations
    d = vit + ½at²
    acceleration = g*sin(Θ)
    Mass = Force/Acceleration


    3. The attempt at a solution

    d = vit + ½at²
    3.2 = ½a(2.2)²
    a = 1.322
    Using the distance equation and solving for acceleration, the acceleration of Peter would be 1.322 m/s²

    F = mass x acceleration
    F = 60 kg x 1.322
    F = 79.338 N
    This is the force of Peter as he falls.

    acceleration of weight m = 9.81*sin(59)
    = 8.408 m/s²

    Mass of m = Force/Acceleration
    = 79.338 N / 8.408 m/s²
    = 9.435 kg

    My final answer was very much incorrect. Is there another formula I was supposed to use?
     
  2. jcsd
  3. Oct 14, 2007 #2
    From what I understand the questions wants a counter mass m that is required to lower peter under the condtions.

    EDIT: If i understood it correctly, I worked it out and got an answer of 51.91kg.

    If this is the correct answer then I will walk you through my solution.
    thanks.
     
    Last edited: Oct 14, 2007
  4. Oct 14, 2007 #3
    Yes. I do believe that is what the problem in asking. I found the mass of the counterweight to be about 9kg. That's too small a weight to work, however.

    EDIT: I checked the answers and 51.91 kg is the correct answer. =)
     
    Last edited: Oct 14, 2007
  5. Oct 14, 2007 #4
    How did you arrive at your solution?
     
  6. Oct 14, 2007 #5
    Ok. So peter is going vertically down so let us convert his mass to weight by gravity which is equal to 9.81*60kg=588.6 N

    You figured out the acceleration of the entire network to be 1.322 m/s^2.

    From this acceleration that we need for the network to work, we need peters force downwards which you calculated to be 79.333 N.

    This is the force that will be pulling peter down and pulling the mass m up the slope.

    If peter's force down is 79.33 N with the desired acceleration, this value will be the NET force required for there to be an acceleration of 1.322 m/s^2.

    So weight of peter due to gravity(588.6N) - weight of mass due to gravity = 79.33. (subtract because they are pulling opposite ways)
    You know peters weight from step 1 and voila you have the weight of the mass m. divide by 9.81 and you have the mass m.

    Do you understand?
     
    Last edited: Oct 14, 2007
  7. Oct 14, 2007 #6
    Yes, I understand it now. Thank you very much!
     
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